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For a statistical data x_1, x_2, ldots, x_10 of 10 values, a student obtained the mean as 5.5 and sum_i=1^10 x_i^2 = 371 . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is :

Solution & Explanation

### Related Formula The standard statistical variance equation for a sample size n is defined as: sigma^2 = fracsum x_i^2n - (barx)^2 ### Core Logic Calculate the initial incorrect \sum of observations using the given incorrect mean: barx_textold = 5.5 = fracsum x_textold10 implies sum x_textold = 55 The given incorrect \sum of squares is: sum x_textold^2 = 371 ### Step 1: Compute Corrected Sum of Observations Adjust the linear \sum by subtracting the incorrect inputs and adding the true values: sum x_textnew = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60 Calculate the new corrected mean value: barx_textnew = frac6010 = 6 ### Step 2: Compute Corrected Sum of Squares Adjust the \sum of squares by swapping the squared entries: sum x_textnew^2 = 371 - (4^2 + 5^2) + (6^2 + 8^2) sum x_textnew^2 = 371 - (16 + 25) + (36 + 64) sum x_textnew^2 = 371 - 41 + 100 = 430 ### Step 3: Calculate Corrected Variance Substitute the corrected values into the standard variance formula: sigma_textnew^2 = fracsum x_textnew^210 - (barx_textnew)^2 sigma_textnew^2 = frac43010 - (6)^2 = 43 - 36 = 7 ### Pattern Recognition When updating statistical aggregates like mean and variance after data correction, always compute the corrected linear \sum and \sum of squares separately before recombining them into the variance formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

Reference Study Guides

Statistics

More Statistics Previous-Year Questions

Q61 2025 Mean and Variance
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to :
  • A. 105
  • B. 103
  • C. 100
  • D. 106

Solution

### Related Formula textMean: barx = fracsum x_iN textVariance: sigma^2 = fracsum x_i^2N - barx^2 ### Core Logic We set up algebraic equations using the definitions of mean and variance to determine the values of a+b and ab. ### Step 1: Apply the Mean condition Given mean barx = 9 for N=8 observations: frac6 + 4 + a + 8 + b + 12 + 10 + 138 = 9 53 + a + b = 72 implies a + b = 19 quad text--- (1) ### Step 2: Apply the Variance condition Given variance sigma^2 = 9.25 = frac374: frac36 + 16 + a^2 + 64 + b^2 + 144 + 100 + 1698 - 81 = frac374 frac529 + a^2 + b^28 = 81 + 9.25 = 90.25 = frac3614 529 + a^2 + b^2 = 722 implies a^2 + b^2 = 193 quad text--- (2) ### Step 3: Solve for ab and calculate the target expression We know (a+b)^2 = a^2 + b^2 + 2ab. Substitute equations (1) and (2): 19^2 = 193 + 2ab 361 = 193 + 2ab implies 2ab = 168 implies ab = 84 Now, calculate the target value: a + b + ab = 19 + 84 = 103 ### Pattern Recognition Direct symmetric evaluation: Statistics problems in JEE with missing observations often ask for symmetric combinations of variables like a+b+ab or a^2+b^2. These can be calculated using quadratic expansions without solving for a and b individually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics
Q62 2025 Mean and Variance
The mean and standard deviation of 100 observations are 40 and 5.1, respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are mu and sigma respectively, then 10(mu + sigma) is equal to
  • A. 445
  • B. 451
  • C. 447
  • D. 449

Solution

### Related Formula Mean command: barx = fracsum x_in Variance command: sigma^2 = fracsum x_i^2n - (barx)^2 ### Core Logic Given values: n = 100, quad barx_textold = 40, quad sigma_textold = 5.1 Incorrect item = 50, Correct replacement = 40. Find the correct mean mu: sum x_i,textold = 100 times 40 = 4000 sum x_i,textcorrect = 4000 - 50 + 40 = 3990 mu = frac3990100 = 39.9 ### Step 1: Compute Correct Variance From the incorrect variance formulation: sigma_textold^2 = (5.1)^2 = 26.01 26.01 = fracsum x_i,textold^2100 - (40)^2 = fracsum x_i,textold^2100 - 1600 fracsum x_i,textold^2100 = 1626.01 implies sum x_i,textold^2 = 162601 Adjust the squared summation block: sum x_i,textcorrect^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701 Now compute the updated variance sigma^2: sigma^2 = frac161701100 - (39.9)^2 sigma^2 = 1617.01 - 1592.01 = 25 sigma = sqrt25 = 5 ### Step 2: Final Calculation Substitute the evaluated correct parameters: 10(mu + sigma) = 10(39.9 + 5) = 10(44.9) = 449 ### Pattern Recognition Notice how the subtraction of 50^2 and subsequent addition of 40^2 directly reduces the total squared variance sum by an exact round value of 900, landing beautifully on a perfect square root target value of 25. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics
Q70 2025 Measures of Dispersion
Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and sqrt2 respectively. Then the mean deviation about the mode of these observations is :
  • A. 1
  • B. frac34
  • C. 2
  • D. frac12

Solution

### Core Logic We have 8 observations: 2, 3, 3, 4, 5, 7, a, b. Given mean barx = 4: frac2 + 3 + 3 + 4 + 5 + 7 + a + b8 = 4 implies 24 + a + b = 32 implies a + b = 8 ### Step 1: Using the Variance Property Given standard deviation sigma = sqrt2 implies textVariance sigma^2 = 2. The variance formula is sigma^2 = fracsum x_i^2n - (barx)^2: 2 = frac2^2 + 3^2 + 3^2 + 4^2 + 5^2 + 7^2 + a^2 + b^28 - 4^2 2 = frac4 + 9 + 9 + 16 + 25 + 49 + a^2 + b^28 - 16 = frac112 + a^2 + b^28 - 16 2 + 16 = frac112 + a^2 + b^28 implies 18 times 8 = 112 + a^2 + b^2 144 = 112 + a^2 + b^2 implies a^2 + b^2 = 32 ### Step 2: Solving for a and b and finding the Mode We know (a+b)^2 = a^2 + b^2 + 2ab implies 8^2 = 32 + 2ab implies 64 = 32 + 2ab implies 2ab = 32 implies ab = 16. Solving a+b=8 and ab=16 gives a=4 and b=4. The complete data set is \2, 3, 3, 4, 4, 4, 5, 7\. The value appearing with highest frequency is 4 (occurs 3 times), so textMode = 4. ### Step 3: Calculating Mean Deviation about Mode Mean Deviation about Mode is: textM.D. = fracsum |x_i - textMode|n = frac|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|8 = frac2 + 1 + 1 + 0 + 0 + 0 + 1 + 38 = frac88 = 1 ### Pattern Recognition When a+b=2sqrtab (here 8 = 2sqrt16), the roots are guaranteed to be equal, meaning a=b. Recognizing this condition instantly avoids full quadratic polynomial substitution steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics
Q68 2025 Variance and Mean Shifts
Let mathrms_1, mathrmx_2, dots, mathrmx_10 be ten observations such that sum_mathrmi=1^10 (mathrmx_mathrmi - 2) = 30 , sum_mathrmi=1^10 (mathrmx_mathrmi - beta)^2 = 98 , beta > 2 . and their variance is frac45 . If mu and sigma^2 are respectively the mean and the variance of 2(mathrmx_1 - 1) + 4beta, 2(mathrmx_2 - 1) + 4beta, dots, 2(mathrmx_10 - 1) + 4beta, then fracbetamusigma^2 is equal to:
  • A. 100
  • B. 110
  • C. 120
  • D. 90

Solution

### Related Formula textVariance sigma^2 = fracsum x_i^2N - (barx)^2 textVariance(ax + b) = a^2 textVariance(x) ### Core Logic From first equation: sum x_i - 20 = 30 implies sum x_i = 50 implies barx = 5. Using Variance formula: frac45 = fracsum x_i^210 - 25 implies fracsum x_i^210 = 25.8 implies sum x_i^2 = 258 ### Step 1: Determine \beta value Expand sum (x_i - beta)^2 = 98: sum x_i^2 - 2beta sum x_i + 10beta^2 = 98 258 - 2beta(50) + 10beta^2 = 98 implies 10beta^2 - 100beta + 160 = 0 beta^2 - 10beta + 16 = 0 implies (beta - 8)(beta - 2) = 0 Since beta > 2, we select beta = 8. ### Step 2: Calculate target mean and variance The simplified new observation expression is 2x_i - 2 + 4(8) = 2x_i + 30. New Mean mu = 2barx + 30 = 2(5) + 30 = 40. New Variance sigma^2 = 2^2 times textOld Variance = 4 times frac45 = frac165. ### Step 3: Evaluate Target Ratio fracbetamusigma^2 = frac8 times 40frac165 = frac320 times 516 = 100 ### Pattern Recognition Adding a constant shift alters only the mean value, while scale factor parameters modify variance quadratically. Recognizing standard linear modification saves significant step calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

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